4.4: Convective Energy Transport

Our approach to the transport of energy by convection will be somewhat different from that for radiation. For radiation, we knew how much energy there was to carry \(-\left[\mathrm{L}(\mathrm{r}) / 4 \pi \mathrm{r}^2\right]\), and we set about finding the temperature gradient required to carry it. For convection, we will anticipate the answer by calculating the amount of energy that a super-adiabatic temperature gradient will carry. For a wide range of parameters thought to prevail in the stellar interior, we shall discover that the adiabatic gradient is adequate to carry all the required energy. But first we must determine the adiabatic temperature gradient.

a. Adiabatic Temperature Gradient

In Chapter 2 [equation \ref{2.4.6}] we defined polytropic change in terms of a specific heat-like quantity C which is equal to the change in heat with respect to temperature. For an adiabatic change, the gas does no work on the surrounding medium, so that C = 0. The polytropic \(\gamma’\) as defined by equation \ref{2.4.8} is \[\gamma^{\prime}=\gamma=\frac{C_P}{C_V}\label{4.3.1}\]

Using equation \ref{2.4.5} and the ideal-gas law, we have \[\gamma=1+\frac{R}{C_V}=1+\frac{N k}{\frac{3}{2} N k}=\frac{5}{3}\label{4.3.2}\]

or \[n=\frac{3}{2}\label{4.3.3}\]

where \(n\) is the appropriate polytropic index for an adiabatic gas.

Now the polytropic equation of state [equation \ref{2.4.1}] and the ideal-gas law guarantee that \[\begin{aligned}
P & =k \rho^{(n+1) / n} \\
T & =\left(\frac{\mu m_h k}{k}\right) \rho^{1 / n}
\end{aligned}\label{4.3.4}\]

Forming the logarithmic derivative of P and T with respect to ρ we get \[\begin{aligned}
& \frac{1}{P} \frac{d P}{d \rho}=\frac{n+1}{n} \rho^{-1} \\
& \frac{1}{T} \frac{d T}{d \rho}=\frac{1}{n} \rho^{-1}
\end{aligned}\label{4.3.5}\]

Dividing these two equations yields \[\frac{T}{P} \frac{d P}{d T}=n+1\label{4.3.6}\]

which is known as the polytropic temperature gradient and for an adiabatic gas is just \[\frac{d \ln P}{d \ln T}=2.5\label{4.3.7}\]

b. Energy Carried by Convection

Imagine a small element of matter rising as a result of being somewhat hotter than its surroundings (see Figure 4.1). We can express the temperature difference between the gas element and its surroundings in terms of the external temperature gradient and the internal temperature gradient experienced by the small element as it rises. We assume that the element is behaving adiabatically, and so this internal gradient is the adiabatic gradient and the temperature difference is \[\delta T=\left.\frac{d T}{d r}\right|_{\mathrm{ad}} \delta r-\left.\frac{d T}{d r}\right|_{\mathrm{ext}} \delta r \equiv \Delta \nabla T \delta r\label{4.3.8}\]

Figure 4.1 shows a schematic representation of a convective element of gas responding to a small temperature difference.

The flux of energy carried by this small convective element will be \[F_{\mathrm{conv}}=(\Delta \nabla T \delta r) \rho C_P v\label{4.3.9}\]

where v is the average velocity of the convective element which we must estimate. The buoyant force experienced by the convective element will be determined solely by the density difference resulting from its slightly elevated temperature and is \[F_b=\frac{G m(r)}{r^2} \delta \rho\label{4.3.10}\]

Since the convective element rises adiabatically, pressure equilibrium will always be maintained during its ascent. Thus we can relate the variation in density to the variation in temperature by setting the variation of the ideal-gas law to zero. \[\delta P=\frac{k T}{\mu m_h} \delta \rho+\frac{k \rho}{\mu m_h} \delta T=0\label{4.3.11}\]

We may use this and equation \ref{4.3.8} to obtain the average buoyancy acting on the convective element. Initially, the buoyancy force is zero since the gradient difference does not produce a significant force until the element has traveled some distance. Thus, we take the average force to be one-half the maximum force, and we get \[\left\langle F_b\right\rangle=-\frac{1}{2} \frac{G M(r)}{r^2} \frac{\rho}{T} \Delta \nabla T \delta r\label{4.3.12}\]

Now the buoyancy force will continuously accelerate the convective element, giving it a kinetic energy of \((1 / 2) \rho \mathrm{v}^2\) which we can use to get an estimate of the convective velocity v. Thus, \[\left\langle F_b\right\rangle \delta r=\frac{1}{2} \rho v^2=\left[\frac{1}{2} \frac{G M(r)}{r^2} \frac{\rho}{T}(\Delta \nabla T \delta r)\right] \delta r\label{4.3.13}\]

which yields a convective velocity of \[v_{\mathrm{conv}} \approx\left[\frac{G M(r)}{T r^2}\right]^{1 / 2}(\Delta \nabla T)^{1 / 2} \delta r\label{4.3.14}\]

We define \[l=2 \delta r\label{4.3.15}\]

This quantity \(l\) is known as the mixing length and is largely a free parameter of this theory of convection from which it takes its name. Typically it is taken to be of the order of a pressure scale height, and fortunately for the theory of stellar interiors, the results are not too sensitive to its exact value. In terms of the mixing length, the convective flux becomes \[F_{\mathrm{conv}}=C_P \rho\left[\frac{G M(r)}{T r^2}\right]^{1 / 2}(\Delta \nabla T)^{3 / 2} \frac{l^2}{4}\label{4.3.16}\]

Now all that remains is to estimate the difference in temperature gradients necessary to transport the energy of the star. We will require that the convection carry all the internal energy flowing through the star, so that \[F_{\mathrm{conv}}=\frac{L(r)}{4 \pi r^2}\label{4.3.17}\]

which yields the gradient difference of \[\Delta \nabla T=\left[\frac{L^2(r) T}{C_P^2 \rho^2 G M(r) \pi^2 l^4 r^2}\right]^{1 / 3}\label{4.3.18}\]

To arrive at some estimate of the significance of this result, let us compare it to the adiabatic gradient. We use the adiabatic temperature gradient in equation \ref{4.3.7}, hydrostatic equilibrium [equation \ref{1.2.28}], and the ideal-gas law to get \[(\nabla T)_{\mathrm{ad}}=\frac{d T}{d P} \frac{d P}{d r}=-\frac{\mu m_h G M(r)}{2.5 k r^2}\label{4.3.19}\]

Dividing equation \ref{4.3.18} by the adiabatic gradient we get \[\frac{\Delta \nabla T}{(\nabla T)_{\mathrm{ad}}}=\frac{5}{2}\left(\frac{L(r)}{C_P \pi \rho}\right)^{2 / 3} T^{1 / 3}\left[\frac{r}{G M(r) l}\right]^{4 / 3}\left(\frac{\mu m_h}{k}\right)^{-1}\label{4.3.20}\]

For the sun, there is some evidence that a mixing length of about one-tenth of a solar radius is not implausible. Picking other values for the sun and trying to maximize equation \ref{4.3.20}, we have the following selection: \[\begin{array}{rlrl}
r & =\frac{R}{10} \approx l & M(r) & =M_{\odot} \\
T & =10^7 \mathrm{~K} & \mu & =0.7 \\
\rho & \approx 1 \mathrm{~g} / \mathrm{cm}^3 & L & =L_{\odot} \\
C_P & =3.5 & \frac{\Delta \nabla T}{(\nabla T)_{\mathbf{a d}}} & \approx 4 \times 10^{-3}
\end{array}\label{4.3.21}\]

Thus, it would seem that the convective gradient will lie within a few tenths of a percent of the adiabatic gradient. This is the source of the statement in Chapter 2 that a polytrope of index 3/2 represents convective stars quite well. Indeed, convection is so efficient that the adiabatic gradient will almost always suffice to describe convective stellar interiors. This is fortunate since the mixing length theory we have discussed here is admittedly rather crude. Unfortunately, this efficiency does not carry over into stellar atmospheres because the convective zones are bounded by the surface of the star, dropping the mixing lengths to numbers comparable to the photon mean free path so that radiation competes effectively with convection regardless of the temperature gradient. For stellar interiors, the photon mean free path is measured in centimeters and the mixing length in fractions of a stellar radius. Thus convection, when established, will always be able to carry the stellar luminosity with a temperature gradient close to the adiabatic gradient.